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<h2 class="hd hd-2 unit-title">Introduction to Maxwell's Equations in Vacuum</h2>
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<p>We now have a complete set of Maxwell's equations, originally developed to describe what happens to stationary and moving charges. Let's remove all charges and see what we are left with - electromagnetic radiation.</p><p>In this lesson, we will assume that we already know the solution for radiation and we will show how it is consistent with Maxwell's Equations in vacuum. The solution we will explore is called <i>plane waves</i>.</p>
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<h2 class="hd hd-2 unit-title">L34v1: Maxwell's Equations in Vacuum</h2>
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<h3 class="hd hd-2">L34v1: Maxwell's Equations in Vacuum</h3>
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<h2 class="hd hd-2 unit-title">L34v2: Plane Wave Solution to Maxwell's Equations in Vacuum</h2>
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<h2 class="hd hd-2 unit-title">L34Q1: Plane Wave Solution</h2>
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Plane Wave Solution
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<h2 class="hd hd-2 unit-title">L34v3: Ampere-Maxwell Equation Applied to the Plane Wave</h2>
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<p>
Consider the planar electromagnetic wave propagating along the [mathjaxinline]+x[/mathjaxinline] direction as shown in the left figure above. The electric field oscillates along the [mathjaxinline]y[/mathjaxinline]-direction, [mathjaxinline]\mathbf{\vec{E}}=E_ y(x,t)\mathbf{\hat{j}}[/mathjaxinline], and the magnetic field along the [mathjaxinline]z[/mathjaxinline] direction, [mathjaxinline]\mathbf{\vec{B}}=B_ z(x,t)\mathbf{\hat{k}}[/mathjaxinline]. We will assume that these fields are a solution of Maxwell's equations in free space. In particular, they are a solution of the Ampere-Maxwell law: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \oint _{\text {loop}} \mathbf{\vec{B}}\cdot d\mathbf{\vec{s}} = \mu _0\epsilon _0\dfrac {d}{dt}\underset {\text {area }}{\Large \iint }\mathbf{\vec{E}} \cdot d\mathbf{\vec{A}}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
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</tr>
</table>
<p>
Note that this version has no &#8220;real" current term since we are assuming that the wave is propagating through vacuum where there are no charges or currents present. </p>
<p>
We will apply the Ampere-Maxwell law along the small rectangular loop of length [mathjaxinline]l[/mathjaxinline] and width [mathjaxinline]\Delta x[/mathjaxinline] contained in the [mathjaxinline](x,z)[/mathjaxinline] - plane as shown in the middle figure above. The direction of circulation is chosen to be counterclockwise as viewed from the [mathjaxinline]+y[/mathjaxinline] axis, and therefore the unit vector perpendicular to the rectangle enclosed by the loop is [mathjaxinline]\mathbf{\hat{n}} = +\mathbf{\hat{j}}[/mathjaxinline] as shown. </p>
<p><b class="bfseries">(Part a)</b> In the right figure, the left and right sides of the rectangle are at [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]x+\Delta x[/mathjaxinline], respectively. Because we are assuming that the wave is planar and moving along the x-axis, the magnetic field is only a function of [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]t[/mathjaxinline]. As a result, the magnetic field has the same value of [mathjaxinline]B_ z(x,t)[/mathjaxinline] at any point along the left side, and the same value of [mathjaxinline]B_ z(x+\Delta x,t)[/mathjaxinline] at any point along the right side. Define [mathjaxinline]\Delta B_ z = B_ z(x+\Delta x,t)-B_ z(x,t)[/mathjaxinline]. Calculate [mathjaxinline]\oint _{\text {loop}} \mathbf{\vec{B}}\cdot d\mathbf{\vec{s}}[/mathjaxinline], the line integral of the magnetic field along the rectangular loop. Express your answer in terms of DeltaB_z for [mathjaxinline]\Delta B_ z[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and Deltax for [mathjaxinline]\Delta x[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\oint _{\text {loop}} \mathbf{\vec{B}}\cdot d\mathbf{\vec{s}}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part b)</b> Consider the width of the loop to be very small, [mathjaxinline]\Delta x \rightarrow 0[/mathjaxinline]. In this limit, we can assume that the electric field throughout the area enclosed by the loop is approximately constant and its component is equal to [mathjaxinline]E_ y(x+\Delta x/2,t)[/mathjaxinline]. Calculate the flux of the electric field through the rectangle enclosed by the loop. Express your answer in terms of E_y for [mathjaxinline]E_ y(x+\Delta x/2,t)[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and Deltax for [mathjaxinline]\Delta x[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\underset {\text {area }}{\Large \iint }\vec{E} \cdot d\vec{A}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part c)</b> Replace the results of parts (a) and (b) in the Ampere-Maxwell law (eq. 1), and after taking the limit of [mathjaxinline]\Delta x \rightarrow 0[/mathjaxinline] we obtain: </p>
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<text> [mathjaxinline]\dfrac {\partial B_ z}{\partial x}=+\mu _0 \epsilon _0\dfrac {\partial E_ y}{\partial t}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {\partial B_ z}{\partial x}=-\mu _0 \epsilon _0\dfrac {\partial E_ y}{\partial t}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {\partial B_ z}{\partial t}=+\mu _0 \epsilon _0\dfrac {\partial E_ y}{\partial x}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {\partial B_ z}{\partial t}=-\mu _0 \epsilon _0\dfrac {\partial E_ y}{\partial x}[/mathjaxinline]</text>
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<h3 class="hd hd-2">L34v3: Ampere-Maxwell Equation Applied to the Plane Wave</h3>
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<h2 class="hd hd-2 unit-title">L34v4: Faraday's Law Applied to the Plane Wave</h2>
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Faraday&#39;s Law and the Wave Equation
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Consider the planar electromagnetic wave propagating along the [mathjaxinline]+x[/mathjaxinline] direction as shown in the left figure above. The electric field oscillates along the [mathjaxinline]y[/mathjaxinline]-direction, [mathjaxinline]\mathbf{\vec{E}}=E_ y(x,t)\mathbf{\hat{j}}[/mathjaxinline], and the magnetic field along the [mathjaxinline]z[/mathjaxinline] direction, [mathjaxinline]\mathbf{\vec{B}}=B_ z(x,t)\mathbf{\hat{k}}[/mathjaxinline]. We will assume that these fields are a solution of Maxwell's equations in free space. In particular, they are a solution of Faraday's law: </p>
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \oint _{\text {loop}} \mathbf{\vec{E}}\cdot d\mathbf{\vec{s}} = -\dfrac {d}{dt}\underset {\text {area }}{\iint \mathbf{\vec{B}} \cdot d\mathbf{\vec{A}}}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
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We will apply Faraday's law along the small rectangular loop of length [mathjaxinline]l[/mathjaxinline] and width [mathjaxinline]\Delta x[/mathjaxinline] contained in the [mathjaxinline](x,y)[/mathjaxinline] - plane as shown in the middle figure above. The direction of circulation is chosen to be counterclockwise as viewed from the [mathjaxinline]+z[/mathjaxinline] axis, and therefore the unit vector perpendicular to the rectangle enclosed by the loop is [mathjaxinline]\mathbf{\hat{n}} = +\mathbf{\hat{k}}[/mathjaxinline] as shown. </p>
<p><b class="bfseries">(Part a)</b> In the right figure, the left and right vertical sides of the rectangle are at [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]x+\Delta x[/mathjaxinline], respectively. Because we are assuming that the wave is planar and moving along the x-axis, the electric field is only a function of [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]t[/mathjaxinline]. As a result, the electric field has the same value of [mathjaxinline]E_ y(x,t)[/mathjaxinline] at any point along the left vertical side, and the same value of [mathjaxinline]E_ y(x+\Delta x,t)[/mathjaxinline] at any point along the right vertical side. Define [mathjaxinline]\Delta E_ y = E_ y(x+\Delta x,t)-E_ y(x,t)[/mathjaxinline]. Calculate [mathjaxinline]\oint _{\text {loop}} \mathbf{\vec{E}}\cdot d\mathbf{\vec{s}}[/mathjaxinline], the line integral of the electric field along the rectangular loop. Express your answer in terms of DeltaE_y for [mathjaxinline]\Delta E_ y[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and Deltax for [mathjaxinline]\Delta x[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\oint _{\text {loop}} \mathbf{\vec{E}}\cdot d\mathbf{\vec{s}}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part b)</b> Consider the width of the loop to be very small, [mathjaxinline]\Delta x \rightarrow 0[/mathjaxinline]. In this limit, we can assume that the magnetic field throughout the area enclosed by the loop is approximately constant and its component is equal to [mathjaxinline]B_ z(x+\Delta x/2,t)[/mathjaxinline]. Calculate the flux of the magnetic field through the rectangle enclosed by the loop. Express your answer in terms of B_z for [mathjaxinline]B_ z(x+\Delta x/2,t)[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and Deltax for [mathjaxinline]\Delta x[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]\underset {\text {area }}{\iint \vec{B} \cdot d\vec{A}}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part c)</b> Replace the results of parts (a) and (b) in Faraday's law (eq. 1), and after taking the limit of [mathjaxinline]\Delta x \rightarrow 0[/mathjaxinline] we obtain: </p>
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w12_24_4_1">
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<text> [mathjaxinline]\dfrac {\partial B_ z}{\partial x}=-\dfrac {\partial E_ y}{\partial t}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {\partial E_ y}{\partial x}=-\dfrac {\partial B_ z}{\partial t}[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L34v5: Wave Equation for E and B</h2>
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Wave Equation for E and B
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<p>
Consider electric and magnetic fields that only depend on [mathjaxinline](x,t)[/mathjaxinline] and are given by [mathjaxinline]\mathbf{\vec{E}}=E_ y(x,t)\mathbf{\hat{j}}[/mathjaxinline] and [mathjaxinline]\mathbf{\vec{B}}=B_ z(x,t)\mathbf{\hat{k}}[/mathjaxinline]. If these fields are solutions of the Faraday and Ampere-Maxwell equations, the components of the fields are related by: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {\partial E_ y}{\partial x }[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle -\dfrac {\partial B_ z}{\partial t}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
<tr id="a0000000004">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {\partial B_ z}{\partial x}[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle -\mu _0\epsilon _0\dfrac {\partial E_ y}{\partial t}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>2</span>)</span>
</td>
</tr>
</table>
<p><b class="bfseries">(Part a)</b> Which of the following expressions will you obtain after taking the partial derivative of (eq. 1) with respect to [mathjaxinline]x[/mathjaxinline], and the partial derivative of (eq. 2) with respect to [mathjaxinline]t[/mathjaxinline]? Check all that apply. </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w12_26_2_1">
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<text>[mathjaxinline]\dfrac {\partial ^2 E_ y}{\partial x^2}=+\dfrac {\partial }{\partial x}(\dfrac {\partial B_ z}{\partial t})[/mathjaxinline]</text>
</label>
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<div class="field">
<input type="checkbox" name="input_checkpoint_w12_26_2_1[]" id="input_checkpoint_w12_26_2_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="checkpoint_w12_26_2_1-choice_1-label" for="input_checkpoint_w12_26_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_26_2_1">
<text>[mathjaxinline]\dfrac {\partial ^2 E_ y}{\partial x^2}=-\dfrac {\partial }{\partial x}(\dfrac {\partial B_ z}{\partial t})[/mathjaxinline]</text>
</label>
</div>
<div class="field">
<input type="checkbox" name="input_checkpoint_w12_26_2_1[]" id="input_checkpoint_w12_26_2_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="checkpoint_w12_26_2_1-choice_2-label" for="input_checkpoint_w12_26_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_26_2_1">
<text>[mathjaxinline]\dfrac {\partial ^2 E_ y}{\partial x^2}=-\dfrac {\partial }{\partial t}(\dfrac {\partial B_ z}{\partial x})[/mathjaxinline]</text>
</label>
</div>
<div class="field">
<input type="checkbox" name="input_checkpoint_w12_26_2_1[]" id="input_checkpoint_w12_26_2_1_choice_3" class="field-input input-checkbox" value="choice_3"/><label id="checkpoint_w12_26_2_1-choice_3-label" for="input_checkpoint_w12_26_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_26_2_1">
<text>[mathjaxinline]\mu _0\epsilon _0\dfrac {\partial ^2 E_ y}{\partial t^2}=-\dfrac {\partial }{\partial x}(\dfrac {\partial B_ z}{\partial t})[/mathjaxinline]</text>
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<text>[mathjaxinline]\mu _0\epsilon _0\dfrac {\partial ^2 E_ y}{\partial x^2}=\dfrac {\partial }{\partial t}(\dfrac {\partial B_ z}{\partial x})[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> Use the results of part (a) to obtain a differential equation including only the electric field component [mathjaxinline]E_ y(x,t)[/mathjaxinline]. </p>
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<text>[mathjaxinline]\dfrac {\partial ^2 E_ y}{\partial x^2}=-\mu _0\epsilon _0\dfrac {\partial ^2 E_ y}{\partial t^2}[/mathjaxinline]</text>
</label>
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<text>[mathjaxinline]\dfrac {\partial ^2 E_ y}{\partial x^2}=+\mu _0\epsilon _0\dfrac {\partial ^2 E_ y}{\partial t^2}[/mathjaxinline]</text>
</label>
</div>
<div class="field">
<input type="checkbox" name="input_checkpoint_w12_26_3_1[]" id="input_checkpoint_w12_26_3_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="checkpoint_w12_26_3_1-choice_2-label" for="input_checkpoint_w12_26_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_26_3_1">
<text>[mathjaxinline]\dfrac {\partial ^2 E_ y}{\partial x^2}=\mu _0\epsilon _0\dfrac {\partial }{\partial t}(\dfrac {\partial E_ y}{\partial x})[/mathjaxinline]</text>
</label>
</div>
<span id="answer_checkpoint_w12_26_3_1"/>
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<p><b class="bfseries">(Part c)</b> Which of the following expressions will you obtain after taking the partial derivative of (eq. 1) with respect to [mathjaxinline]t[/mathjaxinline], and the partial derivative of (eq. 2) with respect to [mathjaxinline]x[/mathjaxinline]? Check all that apply. </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w12_26_4_1">
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<text>[mathjaxinline]\dfrac {\partial ^2 B_ z}{\partial x^2}=+\mu _0\epsilon _0\dfrac {\partial }{\partial t}(\dfrac {\partial E_ y}{\partial x})[/mathjaxinline]</text>
</label>
</div>
<div class="field">
<input type="checkbox" name="input_checkpoint_w12_26_4_1[]" id="input_checkpoint_w12_26_4_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="checkpoint_w12_26_4_1-choice_1-label" for="input_checkpoint_w12_26_4_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_26_4_1">
<text>[mathjaxinline]\dfrac {\partial ^2 B_ z}{\partial x^2}=-\mu _0\epsilon _0\dfrac {\partial }{\partial x}(\dfrac {\partial E_ y}{\partial t})[/mathjaxinline]</text>
</label>
</div>
<div class="field">
<input type="checkbox" name="input_checkpoint_w12_26_4_1[]" id="input_checkpoint_w12_26_4_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="checkpoint_w12_26_4_1-choice_2-label" for="input_checkpoint_w12_26_4_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_26_4_1">
<text>[mathjaxinline]\dfrac {\partial ^2 B_ z}{\partial t^2}=+\dfrac {\partial }{\partial x}(\dfrac {\partial E_ y}{\partial t})[/mathjaxinline]</text>
</label>
</div>
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<text>[mathjaxinline]\dfrac {\partial ^2 B_ z}{\partial t^2}=-\dfrac {\partial }{\partial x}(\dfrac {\partial E_ y}{\partial t})[/mathjaxinline]</text>
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<p><b class="bfseries">(Part d)</b> Use the results of part (c) to obtain a differential equation including only the magnetic field component [mathjaxinline]B_ z(x,t)[/mathjaxinline]. </p>
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<text>[mathjaxinline]\dfrac {\partial ^2 B_ z}{\partial x^2}=-\mu _0\epsilon _0\dfrac {\partial ^2 B_ z}{\partial t^2}[/mathjaxinline]</text>
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<text>[mathjaxinline]\dfrac {\partial ^2 B_ z}{\partial x^2}=+\mu _0\epsilon _0\dfrac {\partial ^2 B_ z}{\partial t^2}[/mathjaxinline]</text>
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<text>[mathjaxinline]\dfrac {\partial }{\partial t}(\dfrac {\partial B_ z}{\partial x})=\mu _0\epsilon _0\dfrac {\partial ^2 B_ z}{\partial x^2}[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L34Q2: Wave Equation and Traveling Waves</h2>
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Wave Equation and Traveling Waves
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The wave equation in one dimension is given by: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {\partial ^2 f}{\partial x^2}=\dfrac {1}{v^2} \dfrac {\partial ^2 f}{\partial t^2}[/mathjaxinline]
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<span>(<span>1</span>)</span>
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where [mathjaxinline]v[/mathjaxinline] is the speed of the wave, and [mathjaxinline]f(x,t)[/mathjaxinline] is the function describing the shape of the wave. This equation describes any type of one dimensional wave. We can classify waves into two groups, mechanical and electromagnetic waves. Mechanical waves need a medium to propagate, for example sound waves, water waves, and waves on a rope. The speed of a mechanical wave depends on the properties of the medium through which is passes (density of the fluid, mass density, tension on the rope, etc.). On the other hand, electromagnetic waves can propagate in free space (vacuum). </p>
<p><b class="bfseries">(Part a)</b> Consider [mathjaxinline]f(x,t) = A\sin (bxt)[/mathjaxinline], where [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] are positive constants. Express the following answers in terms of [mathjaxinline]A[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]t[/mathjaxinline] as needed. </p>
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<p style="display:inline">Calculate [mathjaxinline]\dfrac {\partial ^2 f}{\partial x^2}=[/mathjaxinline] </p>
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<p style="display:inline">Calculate [mathjaxinline]\dfrac {\partial ^2 f}{\partial t^2}=[/mathjaxinline] </p>
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Wave Equation and Traveling Waves
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Replace the results obtained for [mathjaxinline]\dfrac {\partial ^2 f}{\partial x^2}[/mathjaxinline] and [mathjaxinline]\dfrac {\partial ^2 f}{\partial t^2}[/mathjaxinline] in (eq. 1). Is [mathjaxinline]f(x,t) = A\sin (bxt)[/mathjaxinline] a solution of the wave equation? </p>
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Wave Equation and Traveling Waves
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<p><b class="bfseries">(Part b)</b> Consider [mathjaxinline]f(x,t) = A(x-vt)^4[/mathjaxinline], where [mathjaxinline]A[/mathjaxinline] is a positive constant. Express the following answers in terms of [mathjaxinline]A[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]t[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">Calculate [mathjaxinline]\dfrac {\partial ^2 f}{\partial x^2}=[/mathjaxinline] </p>
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<p style="display:inline">Calculate [mathjaxinline]\dfrac {\partial ^2 f}{\partial t^2}=[/mathjaxinline] </p>
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Wave Equation and Traveling Waves
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Replace the results obtained for [mathjaxinline]\dfrac {\partial ^2 f}{\partial x^2}[/mathjaxinline] and [mathjaxinline]\dfrac {\partial ^2 f}{\partial t^2}[/mathjaxinline] in (eq. 1). Is [mathjaxinline]f(x,t) = A(x-vt)^4[/mathjaxinline] a solution of the wave equation? </p>
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<h2 class="hd hd-2 unit-title">L34v6: Traveling Waves and the Speed of Light</h2>
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Wave Equation and Traveling Waves
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<p>
The wave equation in one dimension is given by: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {\partial ^2 f}{\partial x^2}=\dfrac {1}{v^2} \dfrac {\partial ^2 f}{\partial t^2}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
</table>
<p>
where [mathjaxinline]v[/mathjaxinline] is the speed of the wave, and [mathjaxinline]f(x,t)[/mathjaxinline] is the function describing the shape of the wave. You will show that any function [mathjaxinline]f(x,t)=f(x-vt)[/mathjaxinline], i.e. any function of [mathjaxinline]y=x-vt[/mathjaxinline], is a solution of the wave equation (1). </p>
<p>
In fact, any function of the form [mathjaxinline]f(x,t)=f(x \pm vt)[/mathjaxinline] is a solution. We will check this for [mathjaxinline]f(x,t)=f(x-vt)[/mathjaxinline] but the steps are similar for the other sign option. </p>
<p>
In the following calculations note that [mathjaxinline]f[/mathjaxinline] is a <i class="itshape">function of the function</i> [mathjaxinline]y=x-vt[/mathjaxinline], therefore you will be using the chain rule of calculus when calculating the partial derivatives. Express your answers in terms of (df)/(dy) for [mathjaxinline]\dfrac {df}{dy}[/mathjaxinline], (d2f)/(dy2) for [mathjaxinline]\dfrac {d^2f}{dy^2}[/mathjaxinline], and [mathjaxinline]v[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline"><b class="bfseries">(Part a)</b> Calculate [mathjaxinline]\dfrac {\partial ^2 f}{\partial x^2}=[/mathjaxinline] </p>
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<p style="display:inline"><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]\dfrac {\partial ^2 f}{\partial t^2}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Replace the results obtained for [mathjaxinline]\dfrac {\partial ^2 f}{\partial x^2}[/mathjaxinline] and [mathjaxinline]\dfrac {\partial ^2 f}{\partial t^2}[/mathjaxinline] in (eq. 1). Is [mathjaxinline]f(x-vt)[/mathjaxinline] a solution of the wave equation? </p>
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<p><b class="bfseries">(Part d)</b> Consider the one dimensional wave equation obtained in the previous video for the y-component of the electric field: </p>
<table id="a0000000009" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000010">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {\partial ^2 E_ y}{\partial x^2}=\mu _0\epsilon _0 \dfrac {\partial ^2 E_ y}{\partial t^2}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>2</span>)</span>
</td>
</tr>
</table>
<p>
Assume that [mathjaxinline]E_ y[/mathjaxinline] is a given function of [mathjaxinline]x-vt[/mathjaxinline], [mathjaxinline]E_ y=f(x-vt)[/mathjaxinline]. Repeat the steps in part (a) and (b) to calculate the value of [mathjaxinline]v[/mathjaxinline] that makes [mathjaxinline]f(x-vt)[/mathjaxinline] be a solution of (eq. 2). Express your answer in terms of mu_0 for [mathjaxinline]\mu _0[/mathjaxinline] and epsilon_0 for [mathjaxinline]\epsilon _0[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]v=[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">L34Q3: Traveling Waves - Direction of Propagation</h2>
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Traveling Waves - Direction of Propagation
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At time [mathjaxinline]t=0[/mathjaxinline], a traveling pulse has its peak at position [mathjaxinline]x=0[/mathjaxinline] and is described by the function [mathjaxinline]f(x)[/mathjaxinline] as shown in the central figure above. It is known that the pulse is traveling with speed [mathjaxinline]v[/mathjaxinline]. At a later time, </p>
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<text>Figure 1 is a pulse travelling to the left and is described by [mathjaxinline]f(x+vt)[/mathjaxinline]</text>
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<text>Figure 1 is a pulse travelling to the left and is described by [mathjaxinline]f(x-vt)[/mathjaxinline]</text>
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<text>Figure 2 is a pulse travelling to the right and is described by [mathjaxinline]f(x+vt)[/mathjaxinline]</text>
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<text>Figure 2 is a pulse travelling to the right and is described by [mathjaxinline]f(x-vt)[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L34Q4: The Speed of a Pulse</h2>
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The Speed of a Pulse
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Consider a very long rope at rest on a frictionless surface and placed parallel to the [mathjaxinline]x[/mathjaxinline]-axis. If its left end is quickly shaken up and down once, a perturbation of the shape of a pulse propagates along the rope from the left to the right end. If there is no energy losses, the pulse will propagate along the rope without changing its original shape. </p>
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<p><b class="bfseries">(Part a)</b> In the figure above, the displacement in centimeters of an ideal rope with respect to its equilibrium position defined by [mathjaxinline]y=0[/mathjaxinline] is shown. The travelling pulse is shown at different times. The speed of the pulse is </p>
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<p style="display:inline">[mathjaxinline]v =[/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_checkpoint_w12_27_2_1">[mathjaxinline]\mathrm{(m\cdot s^{-1} ) }[/mathjaxinline]</span>
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<p><b class="bfseries">(Part b)</b> If the shape of the pulse at time [mathjaxinline]t=0[/mathjaxinline] can be modelled by the the function [mathjaxinline]y(x) = A e^{-Bx^2}[/mathjaxinline], where [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline] are postive constants, what is [mathjaxinline]y(x,t)[/mathjaxinline], the mathematical representation of the shape of pulse at any instant of time [mathjaxinline]t&gt;0[/mathjaxinline]. Express your answer in terms of [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]t[/mathjaxinline]. </p>
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Wave Number and Angular Frequency
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<p>
Consider an electric field [mathjaxinline]\mathbf{\vec{E}}=E_ y(x,t)\mathbf{\hat{j}}[/mathjaxinline], and a magnetic field, [mathjaxinline]\mathbf{\vec{B}}=B_ z(x,t)\mathbf{\hat{k}}[/mathjaxinline]. These fields are a solution of the wave equations below: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {\partial ^2 E_ y}{\partial x^2}[/mathjaxinline]
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<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle {\mu _0\epsilon _0} \dfrac {\partial ^2 E_ y}{\partial t^2}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
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<span>(<span>1</span>)</span>
</td>
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<tr id="a0000000004">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {\partial ^2 B_ z}{\partial x^2}[/mathjaxinline]
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<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle {\mu _0\epsilon _0} \dfrac {\partial ^2 B_ z}{\partial t^2}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>2</span>)</span>
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</table>
<p><b class="bfseries">(Part a)</b> The y-component of the [mathjaxinline]E[/mathjaxinline]-field is given by: [mathjaxinline]E_ y(x,t)=E_0\sin (k(x-v t))[/mathjaxinline], where [mathjaxinline]E_0[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], and [mathjaxinline]v[/mathjaxinline] are positive constants. Which of the following statements are true? (Check all that apply). </p>
<p>
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<text>Only functions of the form [mathjaxinline]f(x-vt)[/mathjaxinline] are solutions of (eq. 1) and because of the extra [mathjaxinline]k[/mathjaxinline], [mathjaxinline]E_ y=E_0\sin (k(x-v t))[/mathjaxinline] is not of that form.</text>
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<text>[mathjaxinline]E_ y=E_0\sin (k(x-v t))[/mathjaxinline] is a solution of (eq. 1) if and only if [mathjaxinline]kv = 1/\sqrt {\mu _0\epsilon _0}[/mathjaxinline].</text>
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<text>[mathjaxinline]E_ y=E_0\sin (k(x-v t))[/mathjaxinline] is a solution of (eq. 1) if and only if [mathjaxinline]v = 1/\sqrt {\mu _0\epsilon _0}[/mathjaxinline].</text>
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<p><b class="bfseries">(Part b)</b> Consider the electric field given by [mathjaxinline]E_ y(x,t)=E_0\sin (k(x-v t))[/mathjaxinline], where [mathjaxinline]E_0[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] are positive constants of units V/m and 1/m, respectively. [mathjaxinline]E_0[/mathjaxinline] is the amplitude of the electric field, and [mathjaxinline]k[/mathjaxinline] is called the wave number. We will examine the dependence of the electric field on the position by looking at the field at a fixed instant of time. For example, take a picture at time [mathjaxinline]t=0[/mathjaxinline]. At this instant, the electric field is only a function of [mathjaxinline]x[/mathjaxinline], [mathjaxinline]E_ y(x,0)=E_0\sin (kx)[/mathjaxinline]. Its plot is shown below. </p>
<center>
<img src="/assets/courseware/v1/30a18fbe5ef61877196512b0bab185de/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w12_32.svg" width="550"/>
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<p>
The distance between two consecutive maximums or two consecutive minimums is called the wavelength and is noted with the letter [mathjaxinline]\lambda[/mathjaxinline]. Calculate the value of the wave number [mathjaxinline]k[/mathjaxinline]. Express your answer in terms of lambda for [mathjaxinline]\lambda[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]k=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Now we will look at the time dependence of the electric field [mathjaxinline]E_ y(x,t)=E_0\sin (k(x-v t))[/mathjaxinline]. If we fixed the postion, for example, set [mathjaxinline]x=0[/mathjaxinline] the E-field is given by [mathjaxinline]E_ y(0,t)=E_0\sin (-kvt)[/mathjaxinline] or [mathjaxinline]E_ y(0,t)=-E_0\sin (kvt)[/mathjaxinline], where we have used [mathjaxinline]\sin (-\alpha )=-\sin (\alpha )[/mathjaxinline]. The plot of [mathjaxinline]E_ y(0,t)[/mathjaxinline] is shown below. </p>
<center>
<img src="/assets/courseware/v1/3c9b34fd29a874bd5604237d47e5688d/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w12_32b.svg" width="550"/>
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<p>
The distance between two consecutive maximums or two consecutive minimums is the period [mathjaxinline]T[/mathjaxinline]. Calculate the product [mathjaxinline]kv[/mathjaxinline]. Express your answer in terms of [mathjaxinline]T[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]kv=[/mathjaxinline] </p>
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The product [mathjaxinline]kv[/mathjaxinline] has units of 1/s and is the angular frequency of the oscillation of the electric field in time: [mathjaxinline]\omega = kv[/mathjaxinline]. With this definition, the electric field can be written as: </p>
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle E_ y(x,t)=E_0\sin (kx-\omega t)[/mathjaxinline]
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<h2 class="hd hd-2 unit-title">L34: Summary of 1D Traveling Planar EM Waves</h2>
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<p><b>Solution to the Wave Equation</b></p><p> If a function \(f(x,t)\) is a solution of the differential equation:</p>
\[\dfrac{d^2 f}{dx^2} =\dfrac{1}{v^2}\dfrac{d^2 f}{dt^2}\]
<p>then \(f(x,t)\) is a one dimensional wave. If, in addition, \(f(x,t)\) is of the form of \(f(x,t)=f(x\pm vt)\), it is a traveling wave that propagates with speed \(v\) along the \(+x\) direction for \(f(x- vt)\), or along the \(-x\) direction for \(f(x+ vt)\). </p>
<p> An electromagnetic wave with electric and magnetic fields given by: \(\mathbf{\vec{E}}=E_y(x,t)\mathbf{\hat{j}}\), and \(\mathbf{\vec{B}}=B_z(x,t)\mathbf{\hat{k}}\) whose components are given by:</p>
\[\mathbf{\vec{E}}=E_y(x,t)\mathbf{\hat{j}}=E_0\sin(kx\pm\omega t)\mathbf{\hat{j}}\]
\[\mathbf{\vec{B}}=\mp B_z(x,t)\mathbf{\hat{k}}=B_0\sin(kx\pm\omega t)\mathbf{\hat{k}}\]
<p>is a traveling wave propagating along the \(+x\) direction for \(\sin(kx- \omega t)\), and along the \(-x\) direction for \(\sin(kx+ \omega t)\).
</p>
<p>The speed of the wave is:</p>
\[v = \dfrac{\omega}{k} = \dfrac{1}{\sqrt{\mu_0\epsilon_0}}=\dfrac{\lambda}{T}\]
<p> where the wavenumber \(k\) and the wavelength \(\lambda\) are related by \(k=\dfrac{2\pi}{\lambda}\), and the angular frequency \(\omega\) and period \(T\) are related by \(\omega = \dfrac{2\pi}{T}\).
</p>
<p><b>Relationship Between the Direction of E and B</b></p>
<p>So far we have just worked with solutions for which this is true, but you can find explicitly from solving Maxwell's Equations in vacuum that the directions of \(\vec{E}\), \(\vec{B}\), and the direction of propagation must be related:</p>
<p>\(dir{\vec{E}\times\vec{B}} = \) direction of wave propagation.</p>
<p>In the example above, if \(E_0\) and \(B_0\) are both positive, then \(dir{\vec{E}\times\vec{B}}=\mathbf{\hat{j}}\times\mathbf{\hat{k}}=+\mathbf{\hat{i}}\), so the wave must be traveling in the \(+x\)-direction, and so we know that we must have \(\sin(kx- \omega t)\).</p>
<p><b>How EM Waves are Created: Radiation</b></p>
<p>EM waves are produced by accelerating charges. Here is a nice visualization showing this process. The ''information" about an electric field propagates away from the charge at the speed of light, so observe how an accelerating charge makes wiggles in these electric field lines.</p>
<div style="position: relative; width: 300px; height: 197px;"><a href="https://phet.colorado.edu/sims/radiating-charge/radiating-charge_en.html" style="text-decoration: none;" rel="external"><img src="https://phet.colorado.edu/sims/radiating-charge/radiating-charge-600.png" alt="Radiating Charge" style="border: none;" width="300" height="197"/><div style="position: absolute; width: 200px; height: 80px; left: 50px; top: 58px; background-color: #FFF; opacity: 0.6; filter: alpha(opacity = 60);"/><table style="position: absolute; width: 200px; height: 80px; left: 50px; top: 58px;"><tr><td style="text-align: center; color: #000; font-size: 24px; font-family: Arial,sans-serif;">Click to Run</td></tr></table></a></div>
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<h2 class="hd hd-2 unit-title">L34Q6: Properties of EM Waves</h2>
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Properties of EM Waves
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If you have a plane EM wave whose electric field can be described by: </p>
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \vec{\mathbf{E}} =-E_{0} \sin \left[ k \left(-y + v t\right)\right] \hat{\mathbf{i}}[/mathjaxinline]
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<p><b class="bfseries">(Part a)</b> What direction is the wave traveling? </p>
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<text> [mathjaxinline]+\hat{i}[/mathjaxinline]</text>
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<text> [mathjaxinline]+\hat{j}[/mathjaxinline]</text>
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<text> [mathjaxinline]-\hat{j}[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> What is the wavelength of this EM wave? Write your answer in terms of [mathjaxinline]k[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], and [mathjaxinline]c[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\lambda =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> What is the period? <p style="display:inline">[mathjaxinline]T=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_checkpoint_w12_39_4_1" class="text-input-dynamath capa_inputtype inline textline">
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Magnitude of B if you know E
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<p>In any EM wave, there is a fixed ratio between the magnitudes of \(E\) and \(B\). If \(E=20\, V/m\) at some point in the space through which the wave is moving, what is the value of \(B\) at that point? Give your answer in tesla.</p>
<p style="display:inline">[mathjaxinline]B = [/mathjaxinline] </p>
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<p style="display:inline"> (in [mathjaxinline]T[/mathjaxinline])</p>
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<h2 class="hd hd-2 unit-title">L34Q7: Finding E from B</h2>
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Finding E from B
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><p>For the given magnetic field, choose the correct expression for the electric field that satisfies conditions for an electromagnetic plane wave in vacuum:</p>
<p>\(\vec{B}(y,t)=B_{0}\sin{(ky + \omega t)}\hat{k}\)</p>
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